Position vector R of a particle moving x-y plane

AI Thread Summary
The position vector of a particle in the x-y plane is given by r = (2t^3)i + (6-7t^4)j. To find the position, velocity, and acceleration at t = 2s, the displacement can be calculated by substituting t into the equation. The velocity and acceleration can be derived from the first and second derivatives of the position vector, respectively. It is important to remember that r is a vector quantity. The approach to solving the problem is correct, focusing on the derivatives for velocity and acceleration.
chris61986
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Hello, new to these forums! I'm working on my UPI homework, and I just want to verify if I'm doing something correctly. The problem:

The position vector r of a particle moving in the x - y plane given by r = (2t^3)i + (6-7t^4)j, where r is in meters and t is in seconds.
Calculate
r
v
a
when t = 2s

Now, I'm pretty sure I know how to do it, but I don't get a second chance once I turn the homework in! :)

So in the original equation, r is meters. So I'm thinking this is the displacement.
If this is the displacement, I just plug in 2 for t and solve.
Because that's the displacement, I can find a and v from the derivatives of the original function.

Is this correct?
 
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Sounds good. Just remember that \vec r is a vector.
 
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